If $K(G,1)$ and $K(H,1)$ are Eilenberg-MacLane spaces, show that $K(G\ast H,1)$ is also one.

Definition. A path-connected space whose fundamental group is isomorphic to a given group $G$ and which has a contractible universal covering space is called a $\mathbf{K(G,1)}$ space.

For my attempt I will try to construct a "graph of groups" and put a structure on it such that the resulting space has the properties of a $K(G,1)$ space.

Let $\Gamma$ be a connected, directed graph with three vertices associated, respectively, to $G,H$, and $I$ where $I$ is the trivial group, and connect $I$ to $G$ and $H$ with edges associated with group homomorphisms $\varphi_e$.

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Now we can construct a space $K\Gamma$ by associating $K(G,1)$ with vertex $G$, $K(H,1)$ with vertex $H$, and of course the trivial $K(I,1)$ with vertex $I$. Next I propose filling in a mapping cylinder along each edge $e_1,e_2$ which realizes each homomorphism $\varphi_e$. In particular, we would have two mapping cylinders, one glued at either end to $K(I,1)$ and $K(G,1)$ and the other $K(I,1)$ and $K(H,1)$.

Then it is clear to see that our space $K\Gamma$ is a finite CW-complex, and in particular since $K(I,1)$ is trivial, therefore the edge homomorphisms are injections. It follows from Hatcher Theorem 1B.11 that $K\Gamma$ is a $K(G,1)$ space, and as such it also has contractible universal cover by definition.

A standard application of the Van Kampen Theorem then tells us that $\pi_1 (K\Gamma)\cong \pi_1 (K(G,1)) \ast \pi_1 (K(H,1)) \cong G\ast H$.

Hence $K\Gamma$ is really a $K(G\ast H,1)$ space by construction. $\Box$

Is this correct?

  • 1
    $\begingroup$ Your definition and question use the term differently. Your definition defines $K(G,1)$ as a class of spaces, but in your question, $K(G,1)$ is a space. The real problem is $K(G*H,1)$ - it is unclear what that means. Are you saying there exists a $K(G*H,1)$ space if there is a $K(G,1)$ space and a $K(H,1)$ space? $\endgroup$ Commented May 12, 2022 at 17:00
  • $\begingroup$ Yes. By $K(G\ast H,1)$ I mean a path-connected space, with contractible universal covering, which has fundamental group isomorphic to the free product $G\ast H$. The proposition is: Given that two groups $G$ and $H$ are associated to $K(G,1)$ and $K(H,1)$ spaces respectively, prove that there exists a $K(G\ast H,1)$ space. $\endgroup$ Commented May 12, 2022 at 17:16
  • $\begingroup$ There is an analogous example in Hatcher (pp.88), which shows that the product $K(G,1)\times K(H,1)$ is what he calls a $K(G\times H,1)$ space. The general goal is to extend this to the free product instead of direct product. $\endgroup$ Commented May 12, 2022 at 17:21

1 Answer 1


For any group $G$, there exists an $K(G,1)$ space, so the answer to your question in the title is definitely "yes".

Regarding your explicit construction, I assume you were following page 91/92 of Hatcher, where this is exact construction is given as Example 1B.10? The only mistake I can see in your argument is that the CW-complex $K(\Gamma,1)$ certainly need not be finite, since any of the $K(G,1)$ or $K(H,1)$ may already be infinite.


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